" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
U3 = 8 => a + 2b = 8..........(persamaan I)
U6 = 64=>a + 5b = 64........(persamaan II)
eliminasi:
a + 2b =8
a+ 5b = 64 -
-3b =-56
b = 56/3
a + 2b = 8
a + 2(53/3) = 8
a + 106/3 = 8
a = -82/3
U11 = -82/3 + (11-1)56/3
= -82/3 + 560/3
= 498/3
= 166
dah selesai
u6=64
u3=ar²
ar²=8
u6=ar(pangkat)5
ar(pangkat)5=64
ar(pangkat)5=64
ar²=8
r³=8
r=2
ar²=8
a(2²)=8
a(4)=8
a=8/4
a=2
u11=ar(pangkat)10
=2×2(pangkat)10
=2×1024
=2048
gini cara yang benar